unit4 binary tree

8/7/2019 Unit4 Binary Tree

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Trees

Non-Linear Data Structure


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Non-Linear Data Structure

These data structures donot have their

elements in a sequence.

Trees is an example.

Trees are mainly used to represent datacontaining a hierarchical relationshipbetween elements, ex : records, family

trees and table of contents.


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Terminology of Trees The boxes on the tree are called nodes

The nodes immediately below (to the left and right of) a given nodeare called its children

The node immediately above a given node is called its parent

The (unique) node without a parent is called the root

A node with no children is called a leaf Two children of the same parent are said to be siblings

A node can be an ancestor (e.g. grandparent) or a descendant(e.g. great-grandchild)


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Some tree terminology root: node with no parent

A non-empty tree has exactly one root

leaf: node with no children

siblings: two nodes with the same parent.

path: a sequence of nodes n1, n2, , nk such that ni is the parent of

ni+1 for 1 i < k

in other words, a sequence of hops to get from one node to

another

the length of a path is the number of edges in the path, or 1 less

than the number of nodes in it


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Depth and height depth or level: length of the path from root to the current node

(depth of root = 0)

height: length of the longest path from root to any leaf

empty (null) tree's height: -1

single-element tree's height: 0 tree with a root with children: 1


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Tree example


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Trees

Tree nodes contain two or more links

All other data structures we have discussedonly contain one

Binary trees All nodes contain two links None, one, or both of which may be NULL

The root node is the first node in a tree. Each link in the root node refers to a child

A node with no children is called a leaf node


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Binary Trees Special type of tree in which every node or

vertex has either no children, one child ortwo children.

Characteristics :

Every binary tree has a root pointer whichpoints to the start of the tree.

A binary tree can be empty.

It consists of a node called root, a left subtreeand right subtree both of which are binarytrees themselves.


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Examples : Binary Trees

X X X X

Y YZ Z

A

B C

(1) (2) (3) (4)

Root of tree is node having info as X.(1) Only node is root.

(2) Root has left child Y.

(3) Root X has right child Z.

(4) Root X has left child Y and right child Z whichis again a binary tree with its parent as Z andleft child of Z is A, which in turn is parent forleft child B and right child C.


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Properties of Binary Tree A tree with n nodes has exactly (n-1) edges or

branches. In a tree every node except the root has exactly

one parent (and the root node does not have a

parent). There is exactly one path connecting any two

nodes in a tree.

The maximum number of nodes in a binary treeof height K is 2K+1 -1 where K>=0.


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Representation of Binary Tree Linked List

Every node will consists of information, and twopointers left and right pointing to the left and rightchild nodes.

struct node{

int data;struct node *left;

struct node *right;

};

The topmost node or first node is pointed by a rootpointer which will inform the start of the tree. Rest ofthe nodes are attached either to left if less than parentor right if more or equal to parent.


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Diagram of a binary tree

B

A D

C


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Operations on Binary Tree

Searching an existing node.

Inserting a new node.

Deleting an existing node.

Traversing the tree. Preorder

Inorder

Postorder


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Search Process Initialize a search pointer as the root pointer.

All the data is compared with the data stored in eachnode of the tree starting from root node.

If the data to be searched is equal to the data of thenode then print successful search.

Else if the data is less than nodes data move the pointerto left subtree. Else data is more than nodes data movethe pointer to right subtree.

Keep moving in the tree until the data is found or searchpointer comes to NULL in which case it is anunsuccessful search.


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Example used for Search21

18

197

6 9

8 11

14

13

Root


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Example : Search Initialize temp as root pointer which is node having 21.

Loop until temp!=NULL or ITEM found Compare item=9 with the root 21 of the tree, since

9


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Insert Process For node insertion in a binary search tree, initially the

data that is to be inserted is compared with the data ofthe root data.

If the data is found to be greater than or equal to thedata of the root node then the new node is inserted in

the right subtree of the root node, else left subtree. Now the parent of the right or left subtree is considered

and its data is compared with the data of the new node

and the same procedure is repeated again until a NULLis found which will indicate a space when the new nodehas to be attached. Thus finally the new node is madethe appropriate child of this current node.


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Example used for Insert38

14

238

Root

56

8245

18 70

20


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Example : Insert Suppose the ITEM=20 is the data part of the new node

to be inserted in the tree and the root is pointing to thestart of the tree.

Compare ITEM=20 with the root 38 of the tree. Since 20< 38 proceed to the left child of 38, which is 14.

Compare ITEM=20 with 14, since 20 > 14 proceed to theright child of 14, which is 23.

Compare ITEM=20 with 23 since 20 < 23 proceed to the

left child of 23 which is 18. Compare ITEM=20 with 18 since 20 > 18 and 18 does

not have right child, 10 is inserted as the right child of 18.


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Tree traversals: Inorder traversal prints the node values in ascending order

1. Traverse the left subtree with an inorder traversal

2. Process the value in the node (i.e., print the node value)

3. Traverse the right subtree with an inorder traversal

Preorder traversal

1. Process the value in the node

2. Traverse the left subtree with a preorder traversal

3. Traverse the right subtree with a preorder traversal

Postorder traversal

1. Traverse the left subtree with a postorder traversal

2. Traverse the right subtree with a postorder traversal3. Process the value in the node


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Binary tree traversals three common binary tree traversal orderings

(each one begins at the root):

preorder traversal: the current node is processed, then thenode's left subtree is traversed, then the node's right subtree istraversed (CURRENT-LEFT-RIGHT)

in-order traversal: the node's left subtree is traversed, then thecurrent node itself is processed, then the node's right subtree istraversed (LEFT-CURRENT-RIGHT)

postorder traversal: the node's left subtree is traversed, thenthe node's right subtree is traversed, and lastly the current nodeis processed (LEFT-RIGHT-CURRENT)


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Binary tree preorder traversal order: C F T B R K G

The trick: Walk around the outside of the tree and emit a node'svalue when you touch the left side of the node.

"C"

"G""F"

root

"T"

"R""B"

"K"


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Binary tree in-order traversal order: B T R F K C G

The trick: Walk around the outside of the tree and emit a node'svalue when you touch the bottom side of the node.

"C"

"G""F"

root

"T"

"R""B"

"K"


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Binary tree postorder traversal order: B R T K F G C

The trick: Walk around the outside of the tree and emit a node'svalue when you touch the right side of the node

"C"

"G""F"

root

"T"

"R""B"

"K"


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Delete Process There are four possible conditions are to be taken into account :

i. No node in the tree holds the specified data.

ii. The node containing the data has no children.iii. The node containing the data has exactly one child.

iv. The node containing data has two children.

Condition (i) In this case we simply print the message that the data

item is not present in the tree.

Condition (ii) In this case since the node to be deleted has nochildren the memory occupied by this should be freed and eitherthe left link or the right link of the parent of this node should be set

to NULL. Which of these is to be set to NULL depends uponwhether the node being deleted is a left child or right child of itsparent.


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Condition (iii) In this case since the node to be deleted has one child the solution is to

adjust the pointer of the parent of the node to be deleted such that afterdeletion it points to the child of the node being deleted as in the diagram.

21

18

197

6 9

11

10

Root

24

26

25 30

21

18

197

6 11

10

24

26

25 30

Root


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Condition (iv)In this case the node to be deleted has two children. Consider node 9 before deletion. The inorder successor of thenode 9 is node 10. The data of this inorder successor should now be copied into the node to be deleted and a pointershould be set up pointing to the inorder successor (node 10). This inorder successor would always have one or zerochild. It should then be deleted using the same procedure as for deleting one child or a zero child node. Thus, the wholelogic of deleting a node with two children is to locate the inorder successor, copy its data and reduce the problem to a

simple deletion of a node with one or zero child. This is shown in the example.

21

18

197

6 9

11

10

Root

24

26

25 30

8

21

18

197

6 10

11

Root

24

26

25 30

8

Inorder successor of node 9


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Deletion Condition (i) Print message data not found.

Condition (ii) Since no children so free the node& make either parent->left=NULL or parent->right=NULL.

A A

B C

Parent Parent

left right

X X

In the example1,

parent->left==x so free(x) andmake parent->left=NULL.

In the example2,

parent->right==x so free(x) andmake parent->right=NULL.


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Deletion Condition (iii) Adjust parent to point to the child of the deleted node.

I. Node deleted(X) has only left child :

1) If(parent->left==x) parent->left=x->left.

2) If(parent->right==x) parent->right=x->left.

A A

B

Parent Parent

left right

X XB

C C

left

right


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Deletion Condition (iii) Adjust parent to point to the child of the deleted node.

II. Node deleted(X) has only right child :

1) If(parent->left==x) parent->left=x->right.

2) If(parent->right==x) parent->right=x->right.

A A

B

Parent Parent

left right

X XB

C Cright

right


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Deletion Condition (iv) solution more complex as X has two children, Find inorder successor of the node X. Inorder successor of any node will be first go to right of the

node and then from that node keep going left until NULL encountered, that will be the inordersuccessor.

Copy data of inorder successor to node Xs data. Place pointer at the inorder successor node. Now the inorder succesor will have zero or one child

only (so the complex problem is reduced to condition (iii)). Delete the inorder successor node by using logic of condition (ii) if no children or condition (iiii) if one

right child.

A

B

Parent

left

X

D

rightleft

C

E

left

A

E

Parent

left

X

D

rightleft

C

E

leftHere inorder successor E hasno children condition (ii)


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DeletionA

B

Parent

left

X

D

rightleft

C

E

left

F

right

G H

Inorder

Successor

left right

A

E

Parent

left

X

D

rightleft

C

E

left

F

right

G H

Inorder

Successor

left right

A

E

Parent

left

D

rightleft

C

F

G H

left right

Here inorder successor E hasone right child condition (iii)

If(parent->left==x)

parent->left=x->right.

left


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Application of Tree Expression Tree

Game Tree

unit4 binary tree

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